Let $p,$ $q,$ $r,$ $s$ be real numbers such that $p  +q + r + s = 8$ and
\[pq + pr + ps + qr + qs + rs = 12.\]Find the largest possible value of $s.$
Answer: Squaring the equation $p + q + r + s = 8,$ we get
\[p^2 + q^2 + r^2 + s^2 + 2(pq + pr + ps + qr + qs + rs) = 64.\]Hence, $p^2 + q^2 + r^2 + s^2 = 64 - 2 \cdot 12 = 40.$

By Cauchy-Schwarz,
\[(1^2 + 1^2 + 1^2)(p^2 + q^2 + r^2) \ge (p + q + r)^2.\]Then $3(40 - s^2) \ge (8 - s)^2.$  Expanding, we get $120 - 3s^2 \ge 64 - 16s + s^2,$ so $4s^2 - 16s - 56 \le 0.$  Dividing by 4, we get $s^2 - 4s - 14 \le 0.$  By the quadratic formula, the roots of the corresponding equation $x^2 - 4x - 14 = 0$ are
\[x = 2 \pm 3 \sqrt{2},\]so $s \le 2 + 3 \sqrt{2}.$

Equality occurs when $p = q = r = 2 - \sqrt{2},$ so the maximum value of $s$ is $\boxed{2 + 3 \sqrt{2}}.$